In the previous section we learned how to determine the nth term of an AP. We also saw some solved examples. In this section we will see a few more solved examples.
Solved example 26.15
Find the number of terms in the following APs:
(i) 7, 13, 19, . . . , 205
Solution:
1. This is an AP with a = 7 and d = 6. Because every term can be obtained by adding 6 to the preceding term.
3. Let n be the total number of terms. We have to find the n value of the last term 205
4. We have: nth term = a + (n-1)d ⇒ 205 = 7 + (n-1) ×6
⇒ 205 = 7 + 6n - 6 ⇒ 6n = 205 - 7 + 6 ⇒ 6n = 204 ⇒ n = 34
• So there are a total of 34 terms in the given AP.
(ii) 18, 151⁄2, 13, . . . , -47
Solution:
1. This is an AP with a = 18 and d = -21⁄2. = -5⁄2. Because, every term can be obtained by adding (-21⁄2) to the preceding term.
3. Let n be the total number of terms. We have to find the n value of the last term -47
4. We have: nth term = a + (n-1)d ⇒ -47 = 18 + (n-1) ×-5⁄2
⇒ -47 = 18 -5n⁄2 + 5⁄2 ⇒ 5n⁄2 = 47 + 18 +5⁄2 ⇒ 5n = 94 + 36 + 5 ⇒ 5n = 135 ⇒ n = 27
• So there are a total of 27 terms in the given AP.
Solved example 26.16
Check whether -150 is a term in the AP: 11, 8, 5, 2, . . .
Solution:
1. 1st term a = 11
2. Common difference d = 2 - 5 = -3
3. We have: nth term = a + (n-1)d
4. Let -150 be the nth term. Then substituting the values in (3) we get:
5. -150 = 11 + (n-1) × -3 ⇒ -150 = 11 - 3n + 3 ⇒ 3n = 150 + 11 +3 ⇒ 3n = 164 ⇒ n = 164⁄3
• 'n' is a number. It shows a position. Like 1st, 2nd, 3rd etc., So 'n' should be a whole number.
• 164⁄3 is not a whole number. So -150 is not a term in the given AP.
Solved example 26.17
Determine the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. Substituting the values we get:
(i) 38 = a + (11-1) × d ⇒ 38 = a + 10d
(ii) 73 = a + (16-1) × d ⇒ 73 = a + 15d
3. So we have two equations in two variables a and d. We can easily solve them as follows (Details here):
4. From (i) we get a = 38 - 10d
5. Substituting this in (ii) we get: 73 = (38 - 10d) + 15d
⇒ 73 = 38 + 5d ⇒ 5d = 35 ⇒ d = 7
6. Substituting this value of d in (4) we get: a = 38 - 10 × 7 ⇒ a = 38 -70 ⇒ a = -32
7. So we have a and d. We can now find any term of the AP.
• we have: nth term = a + (n-1)d
• So a31 = -32 + (31-1) ×7 = -32 + 30 × 7 = -32 + 210 = 178
Solved example 26.18
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. Substituting the values we get:
(i) 4 = a + (3-1) × d ⇒ 4 = a + 2d
(ii) -8 = a + (9-1) × d ⇒ -8 = a + 8d
3. So we have two equations in two variables a and d. We can easily solve them as follows (Details here):
4. From (i) we get a = 4 - 2d
5. Substituting this in (ii) we get: -8 = (4 - 2d) + 8d
⇒ -8 = 4 + 6d ⇒ 6d = -12 ⇒ d = -2
6. Substituting this value of d in (4) we get: a = 4 - 2 × (-2) ⇒ a = 4 + 4 ⇒ a = 8
7. So we have a and d. We can now find any term of the AP.
• we have: nth term = a + (n-1)d
• So 0 = 8 + (n-1) ×-2 ⇒ 0 = 8 -2n + 2 ⇒ 2n = 10 ⇒ n = 5
• Thus zero is the 5th term of the AP
Solved example 26.19
The 17th term of an AP exceeds its 10th term by 7. Find the common difference
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. Substituting the values we get:
(i) a17 = a + (17-1) × d ⇒ a17 = a + 16d
(ii) a10 = a + (10-1) × d ⇒ a10 = a + 9d
3. Given that: a10 + 7 = a17. Substituting this in (2) we get:
4. a + 9d + 7 = a + 16d ⇒ 9d + 7 = 16d ⇒ 7d = 7 ⇒ d = 1
Solved example 26.20
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. So 54th term = a54 = a + (54-1) × d ⇒ a54 = a + 53d
3. Let an be the required term. We have: an = a + (n-1)d
4. According to the given condition, an = a54 + 132
5. Substituting (2) and (3) in (4) we get:
a + (n-1)d = a + 53d + 132 ⇒ (n-1)d = 53d + 132
⇒ nd - d = 53d + 132 ⇒ (n-1-53)d = 132 ⇒ (n-54)d = 132
6. But from the given AP, d = 39-27 = 12
7. So from (5) we get (n-54) × 12 = 132 ⇒ n-54 = 132⁄12 = 11 ⇒ n = 11 + 54 = 65
Solved example 26.21
Two APs have the same common difference. The difference between 100th terms is 100, what is the difference between their 1000th terms?
Solution:
1. The two APs have the same common difference. Let it be d
2. Let the 1st terms be a1 and b1 respectively
3. Then for the first AP: a100 = a1 + 99d
4. For the second AP: b100 = b1 + 99d
5. Given that the difference of the 100th term is 100. So we can write:
(a1 + 99d) - (b1 + 99d) = 100 ⇒ a1 + 99d - b1 - 99d = 100 ⇒ a1 - b1 = 100
6. Now we write the 1000th terms:
7. For the first AP: a1000 = a1 + 999d
8. For the second AP: b1000 = b1 + 999d
9. Difference = (a1 + 999d) - (b1 + 999d) = a1 - b1.
10. But from (5) we have: a1 - b1 = 100
11. So the difference between their 1000th terms = 100
Solved example 26.22
How many three-digit numbers are divisible by 7?
Solution:
• There are so many 3 digit numbers. The list is very long but not endless. All the 3 digit numbers fall between 100 and 999 both of them inclusive.
• Out of them, many are divisible by 7. In this problem, we do not have to actually pick out all those which are divisible by 7.
• All we need is the 'number 'of 3-digit numbers which are divisible by 7. Let us try to find it:
1. The first 3-digit number is 100. It is not divisible by 7
• If we divide 100 by 7, we get a reminder 2. [100 = (14 × 7) + 2 = 98 + 2]
• So the '2' is causing the problem. If this '2' was not there, the number would be divisible by 7
• To avoid '2' we can take (100-2) = 98. It is divisible by 7.
• But 98 is a 2-digit number. It falls outside our required range
• Then how do we avoid the problem causing '2'?
• Let us make this '2' into a '7'.
• When a number is divided by '7', if the remainder is '7', then it means that there is no remainder,
• So to get '7' as remainder instead of '2', we must add (7-2) = 5 to the original number
• Thus the next number above 100 which is divisible by 7 is 100 + (7-2) = 100 + 5 = 105
2. The last 3-digit number is 999. It is not divisible by 7
• If we divide 999 by 7, we get a reminder 5. [999 = (142 × 7) + 5 = 994 + 5]
• So the '5' is causing the problem. If this '5' was not there, the number would be divisible by 7
• To avoid '5' we can take (999-5) = 994. It is divisible by 7.
Note: We could make remainder '5' into remainder '7' by adding (7-5) = 2 to 999
That is we take (2+999) = 1001. It is divisible by 7. (1001 ÷ 7 = 143)
But 1001 is a 4-digit number. It falls outside our required range
• So the number below 999 which is divisible by 7 is 999 - 5 = 994
The above results can be seen in a tabular form here.
3. Thus the 1st term of the AP is 105 and the last term is 994. The common difference d is 7
4. So we can write: nth term = a + (n-1)d ⇒ nth term = 105 + (n-1) ×7
6. Let 994 be the nth term. Then we can write:
994 = 105 + (n-1) ×7 ⇒ 994 = 105 + 7n -7 ⇒ 7n = 994 - 98 ⇒ 7n = 896 ⇒ n = 128
7. So the last term 994 is the 128th term. We can write:
There are 128 three-digit numbers which are divisible by 7
Solved example 26.23
How many multiples of 4 lie between 10 and 250?
Solution:
1. The first number given is 10. It is not divisible by 4
• If we divide 10 by 4, we get a reminder 2. [10 = (4 × 2) + 2 = 8 + 2]
• So the next number above 10 which is divisible by 2 is 10 + (4-2) = 10 + 2 = 12
2. The last number given is 250. It is not divisible by 4
• If we divide 250 by 4, we get a reminder 2. [250 = (4 × 62) + 2 = 248 + 2]
• So the number below 250 which is divisible by 2 is 250 - 2 = 248
3. Thus the 1st term of the AP is 12 and the last term is 248. The common difference d is 4
4. So we can write: nth term = a + (n-1)d ⇒ nth term = 12 + (n-1) ×4
6. Let 248 be the nth term. Then we can write:
248 = 12 + (n-1) ×4 ⇒ 248 = 12 + 4n -4 ⇒ 4n = 248 - 8 ⇒ 4n = 240 ⇒ n = 60
7. So the last term 248 is the 60th term. We can write:
• There are 60 numbers between 10 and 250,which are divisible by 4. This is same as:
• There are 60 multiples of 4 between 10 and 250
Solved example 26.24
For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
Solution:
1. We can write:
• nth term of the first AP = 63 + (n-1) × 2 = 61 + 2n [∵ d = 67-65 = 2]
• nth term of the second AP = 3 + (n-1) × 7 = 7n - 4 [∵ d = 17-10 = 7]
2. If these two terms are equal, we can write:
61 + 2n = 7n -4 ⇒ 5n = 65 ⇒ n = 13
Solved example 26.25
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
1. 4th term = a + 3d, 8th term = a + 7d
Sum = 2a + 10d = 24 ⇒ a + 5d = 12
2. 6th term = a + 5d, 10th term = a + 9d
Sum = 2a + 14d = 44 ⇒ a + 7d = 22
3. From (1) we get: a = 12-5d. Substituting this in (2) we get:
12 - 5d + 7d = 22 ⇒ 2d = 10 ⇒ d = 5
4. Substituting this value of d in (1) we get:
a + 5 × 5 = 12 ⇒ a + 25 = 12 ⇒ a = -13
5. So we have a and d. We can now form the AP.
1st term = a1 = a = -13
2nd term = a2 = a + d = -13 + 5 = -8
3rd term = a3 = a + 2d = -13 + 2×5 = -13 + 10 = -3
Thus the AP is: -13, -8, -3, . . .
In the next section we will sum of first n terms of an AP.
Solved example 26.15
Find the number of terms in the following APs:
(i) 7, 13, 19, . . . , 205
Solution:
1. This is an AP with a = 7 and d = 6. Because every term can be obtained by adding 6 to the preceding term.
3. Let n be the total number of terms. We have to find the n value of the last term 205
4. We have: nth term = a + (n-1)d ⇒ 205 = 7 + (n-1) ×6
⇒ 205 = 7 + 6n - 6 ⇒ 6n = 205 - 7 + 6 ⇒ 6n = 204 ⇒ n = 34
• So there are a total of 34 terms in the given AP.
(ii) 18, 151⁄2, 13, . . . , -47
Solution:
1. This is an AP with a = 18 and d = -21⁄2. = -5⁄2. Because, every term can be obtained by adding (-21⁄2) to the preceding term.
3. Let n be the total number of terms. We have to find the n value of the last term -47
4. We have: nth term = a + (n-1)d ⇒ -47 = 18 + (n-1) ×-5⁄2
⇒ -47 = 18 -5n⁄2 + 5⁄2 ⇒ 5n⁄2 = 47 + 18 +5⁄2 ⇒ 5n = 94 + 36 + 5 ⇒ 5n = 135 ⇒ n = 27
• So there are a total of 27 terms in the given AP.
Solved example 26.16
Check whether -150 is a term in the AP: 11, 8, 5, 2, . . .
Solution:
1. 1st term a = 11
2. Common difference d = 2 - 5 = -3
3. We have: nth term = a + (n-1)d
4. Let -150 be the nth term. Then substituting the values in (3) we get:
5. -150 = 11 + (n-1) × -3 ⇒ -150 = 11 - 3n + 3 ⇒ 3n = 150 + 11 +3 ⇒ 3n = 164 ⇒ n = 164⁄3
• 'n' is a number. It shows a position. Like 1st, 2nd, 3rd etc., So 'n' should be a whole number.
• 164⁄3 is not a whole number. So -150 is not a term in the given AP.
Solved example 26.17
Determine the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. Substituting the values we get:
(i) 38 = a + (11-1) × d ⇒ 38 = a + 10d
(ii) 73 = a + (16-1) × d ⇒ 73 = a + 15d
3. So we have two equations in two variables a and d. We can easily solve them as follows (Details here):
4. From (i) we get a = 38 - 10d
5. Substituting this in (ii) we get: 73 = (38 - 10d) + 15d
⇒ 73 = 38 + 5d ⇒ 5d = 35 ⇒ d = 7
6. Substituting this value of d in (4) we get: a = 38 - 10 × 7 ⇒ a = 38 -70 ⇒ a = -32
7. So we have a and d. We can now find any term of the AP.
• we have: nth term = a + (n-1)d
• So a31 = -32 + (31-1) ×7 = -32 + 30 × 7 = -32 + 210 = 178
Solved example 26.18
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. Substituting the values we get:
(i) 4 = a + (3-1) × d ⇒ 4 = a + 2d
(ii) -8 = a + (9-1) × d ⇒ -8 = a + 8d
3. So we have two equations in two variables a and d. We can easily solve them as follows (Details here):
4. From (i) we get a = 4 - 2d
5. Substituting this in (ii) we get: -8 = (4 - 2d) + 8d
⇒ -8 = 4 + 6d ⇒ 6d = -12 ⇒ d = -2
6. Substituting this value of d in (4) we get: a = 4 - 2 × (-2) ⇒ a = 4 + 4 ⇒ a = 8
7. So we have a and d. We can now find any term of the AP.
• we have: nth term = a + (n-1)d
• So 0 = 8 + (n-1) ×-2 ⇒ 0 = 8 -2n + 2 ⇒ 2n = 10 ⇒ n = 5
• Thus zero is the 5th term of the AP
Solved example 26.19
The 17th term of an AP exceeds its 10th term by 7. Find the common difference
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. Substituting the values we get:
(i) a17 = a + (17-1) × d ⇒ a17 = a + 16d
(ii) a10 = a + (10-1) × d ⇒ a10 = a + 9d
3. Given that: a10 + 7 = a17. Substituting this in (2) we get:
4. a + 9d + 7 = a + 16d ⇒ 9d + 7 = 16d ⇒ 7d = 7 ⇒ d = 1
Solved example 26.20
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Solution:
1. Let a be the 1st term and d the common difference
Then we have: nth term = a + (n-1)d
2. So 54th term = a54 = a + (54-1) × d ⇒ a54 = a + 53d
3. Let an be the required term. We have: an = a + (n-1)d
4. According to the given condition, an = a54 + 132
5. Substituting (2) and (3) in (4) we get:
a + (n-1)d = a + 53d + 132 ⇒ (n-1)d = 53d + 132
⇒ nd - d = 53d + 132 ⇒ (n-1-53)d = 132 ⇒ (n-54)d = 132
6. But from the given AP, d = 39-27 = 12
7. So from (5) we get (n-54) × 12 = 132 ⇒ n-54 = 132⁄12 = 11 ⇒ n = 11 + 54 = 65
Solved example 26.21
Two APs have the same common difference. The difference between 100th terms is 100, what is the difference between their 1000th terms?
Solution:
1. The two APs have the same common difference. Let it be d
2. Let the 1st terms be a1 and b1 respectively
3. Then for the first AP: a100 = a1 + 99d
4. For the second AP: b100 = b1 + 99d
5. Given that the difference of the 100th term is 100. So we can write:
(a1 + 99d) - (b1 + 99d) = 100 ⇒ a1 + 99d - b1 - 99d = 100 ⇒ a1 - b1 = 100
6. Now we write the 1000th terms:
7. For the first AP: a1000 = a1 + 999d
8. For the second AP: b1000 = b1 + 999d
9. Difference = (a1 + 999d) - (b1 + 999d) = a1 - b1.
10. But from (5) we have: a1 - b1 = 100
11. So the difference between their 1000th terms = 100
Solved example 26.22
How many three-digit numbers are divisible by 7?
Solution:
• There are so many 3 digit numbers. The list is very long but not endless. All the 3 digit numbers fall between 100 and 999 both of them inclusive.
• Out of them, many are divisible by 7. In this problem, we do not have to actually pick out all those which are divisible by 7.
• All we need is the 'number 'of 3-digit numbers which are divisible by 7. Let us try to find it:
1. The first 3-digit number is 100. It is not divisible by 7
• If we divide 100 by 7, we get a reminder 2. [100 = (14 × 7) + 2 = 98 + 2]
• So the '2' is causing the problem. If this '2' was not there, the number would be divisible by 7
• To avoid '2' we can take (100-2) = 98. It is divisible by 7.
• But 98 is a 2-digit number. It falls outside our required range
• Then how do we avoid the problem causing '2'?
• Let us make this '2' into a '7'.
• When a number is divided by '7', if the remainder is '7', then it means that there is no remainder,
• So to get '7' as remainder instead of '2', we must add (7-2) = 5 to the original number
• Thus the next number above 100 which is divisible by 7 is 100 + (7-2) = 100 + 5 = 105
2. The last 3-digit number is 999. It is not divisible by 7
• If we divide 999 by 7, we get a reminder 5. [999 = (142 × 7) + 5 = 994 + 5]
• So the '5' is causing the problem. If this '5' was not there, the number would be divisible by 7
• To avoid '5' we can take (999-5) = 994. It is divisible by 7.
Note: We could make remainder '5' into remainder '7' by adding (7-5) = 2 to 999
That is we take (2+999) = 1001. It is divisible by 7. (1001 ÷ 7 = 143)
But 1001 is a 4-digit number. It falls outside our required range
• So the number below 999 which is divisible by 7 is 999 - 5 = 994
The above results can be seen in a tabular form here.
3. Thus the 1st term of the AP is 105 and the last term is 994. The common difference d is 7
4. So we can write: nth term = a + (n-1)d ⇒ nth term = 105 + (n-1) ×7
6. Let 994 be the nth term. Then we can write:
994 = 105 + (n-1) ×7 ⇒ 994 = 105 + 7n -7 ⇒ 7n = 994 - 98 ⇒ 7n = 896 ⇒ n = 128
7. So the last term 994 is the 128th term. We can write:
There are 128 three-digit numbers which are divisible by 7
Solved example 26.23
How many multiples of 4 lie between 10 and 250?
Solution:
1. The first number given is 10. It is not divisible by 4
• If we divide 10 by 4, we get a reminder 2. [10 = (4 × 2) + 2 = 8 + 2]
• So the next number above 10 which is divisible by 2 is 10 + (4-2) = 10 + 2 = 12
2. The last number given is 250. It is not divisible by 4
• If we divide 250 by 4, we get a reminder 2. [250 = (4 × 62) + 2 = 248 + 2]
• So the number below 250 which is divisible by 2 is 250 - 2 = 248
3. Thus the 1st term of the AP is 12 and the last term is 248. The common difference d is 4
4. So we can write: nth term = a + (n-1)d ⇒ nth term = 12 + (n-1) ×4
6. Let 248 be the nth term. Then we can write:
248 = 12 + (n-1) ×4 ⇒ 248 = 12 + 4n -4 ⇒ 4n = 248 - 8 ⇒ 4n = 240 ⇒ n = 60
7. So the last term 248 is the 60th term. We can write:
• There are 60 numbers between 10 and 250,which are divisible by 4. This is same as:
• There are 60 multiples of 4 between 10 and 250
Solved example 26.24
For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
Solution:
1. We can write:
• nth term of the first AP = 63 + (n-1) × 2 = 61 + 2n [∵ d = 67-65 = 2]
• nth term of the second AP = 3 + (n-1) × 7 = 7n - 4 [∵ d = 17-10 = 7]
2. If these two terms are equal, we can write:
61 + 2n = 7n -4 ⇒ 5n = 65 ⇒ n = 13
Solved example 26.25
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
1. 4th term = a + 3d, 8th term = a + 7d
Sum = 2a + 10d = 24 ⇒ a + 5d = 12
2. 6th term = a + 5d, 10th term = a + 9d
Sum = 2a + 14d = 44 ⇒ a + 7d = 22
3. From (1) we get: a = 12-5d. Substituting this in (2) we get:
12 - 5d + 7d = 22 ⇒ 2d = 10 ⇒ d = 5
4. Substituting this value of d in (1) we get:
a + 5 × 5 = 12 ⇒ a + 25 = 12 ⇒ a = -13
5. So we have a and d. We can now form the AP.
1st term = a1 = a = -13
2nd term = a2 = a + d = -13 + 5 = -8
3rd term = a3 = a + 2d = -13 + 2×5 = -13 + 10 = -3
Thus the AP is: -13, -8, -3, . . .
No comments:
Post a Comment